Question

What is the difference between Mean and Standard Deviation?

Answer 1

In this type of question we have to consider the concept of mean and standard deviation and then we have to consider the difference between them. We know that in maths mean and standard deviation plays an important role in measurement. We know that the average of all given values of all observations is considered as mean while the measure of distribution of the data is called the standard deviation.

Complete step-by-step solution:
Now we have to find the difference between mean and standard deviation.
The mean of a series of data is the value equal to the sum of the values of all the observations divided by the total number of observations. It is the most commonly used quantifier of central tendency. Also it is easy to measure. The formula for calculation of mean is as follows:
$\Rightarrow \overline{x}=\dfrac{\sum{x}}{n}$
Here, $\sum{x}$ represents the sum of all values of the observations, $n$ represents the total number of observations and $\overline{x}$ represents the mean.
In other words we can also write formula for mean as
$\Rightarrow \overline{x}=\dfrac{\text{Sum of all values of the observations}}{\text{Total number of observations}}$
The standard deviation is considered as the measure of the dispersion of the data values from the mean. The standard deviation measures the absolute variability of the distribution of the data. The symbol $''\sigma ''$ represents the standard deviation. The formula to calculate standard deviation of the given data is:
$\Rightarrow \text{Standard Deviation = }\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n}}$
Here, ${{x}_{i}}$ represents the different values of the observations, $n$ represents the total number of observations and $\overline{x}$ represents the mean.
For Example: Find the mean and standard deviation for $6,2,5,4,12,7$.
As we have to find the mean and standard deviation, we first calculate the mean as follows:
$\Rightarrow Mean=\dfrac{\text{Sum of all values of the observations}}{\text{Total number of observations}}$

& \Rightarrow \overline{x}=\dfrac{\sum{x}}{n} \\\ & \Rightarrow \overline{x}=\dfrac{6+2+5+4+12+7}{6} \\\ & \Rightarrow \overline{x}=\dfrac{36}{6} \\\ & \Rightarrow \overline{x}=6 \\\ \end{aligned}$$ Now, we have to find the standard deviation. For this we have to use the formula $$\Rightarrow \text{Standard Deviation = }\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n}}$$ Now, first we will calculate the square of the sum of the deviation from the mean i.e. $$\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}$$ $$\begin{aligned} & \Rightarrow \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}=\sum\limits_{i=1}^{6}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}={{\left( 6-6 \right)}^{2}}+{{\left( 2-6 \right)}^{2}}+{{\left( 5-6 \right)}^{2}}+{{\left( 4-6 \right)}^{2}}+{{\left( 12-6 \right)}^{2}}+{{\left( 7-6 \right)}^{2}} \\\ & \Rightarrow \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}={{\left( 0 \right)}^{2}}+{{\left( -4 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( 6 \right)}^{2}}+{{\left( 1 \right)}^{2}} \\\ & \Rightarrow \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}=16+1+4+36+1 \\\ & \Rightarrow \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}=58 \\\ \end{aligned}$$ Hence, we will get the standard deviation as $$\begin{aligned} & \Rightarrow \text{Standard Deviation = }\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n}} \\\ & \Rightarrow \text{Standard Deviation = }\sigma =\sqrt{\dfrac{58}{6}} \\\ & \Rightarrow \text{Standard Deviation = }\sigma =3.11 \\\ \end{aligned}$$ **Note:** In this type of question students have to remember the definition as well as formulas for mean and standard deviation. Students have to note that the mean will give the average of the observation while standard deviation gives the deviation of all values of the observation from the mean. Also students have to remember that the square of standard deviation will give them variance of the observation which shows how far a data point has dispersed.